Answer :
To solve this problem, we want to find out how far above the ground the hammer was when it was dropped. We use the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( v \)[/tex] is the speed of the hammer when it hits the floor (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]\( h \)[/tex] is the height above the ground from which the hammer was dropped.
We need to solve for [tex]\( h \)[/tex]. Here's how you do it step-by-step:
1. Square Both Sides to Eliminate the Square Root:
Start with the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]
Square both sides:
[tex]\[ v^2 = 2gh \][/tex]
2. Rearrange the Equation to Solve for [tex]\( h \)[/tex]:
Divide both sides by [tex]\( 2g \)[/tex] to isolate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
3. Substitute the Given Values:
Plug in the values for [tex]\( v \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[ h = \frac{4^2}{2 \times 32} \][/tex]
4. Calculate [tex]\( h \)[/tex]:
Simplify the equation:
- [tex]\( v^2 = 4^2 = 16 \)[/tex]
- [tex]\( 2 \times 32 = 64 \)[/tex]
Now compute:
[tex]\[ h = \frac{16}{64} \][/tex]
5. Simplify the Fraction:
[tex]\[ h = \frac{1}{4} = 0.25 \][/tex]
So, the hammer was dropped from a height of 0.25 feet above the ground. The correct answer is:
B. 0.25 feet
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( v \)[/tex] is the speed of the hammer when it hits the floor (4 feet per second),
- [tex]\( g \)[/tex] is the acceleration due to gravity (32 feet/second²),
- [tex]\( h \)[/tex] is the height above the ground from which the hammer was dropped.
We need to solve for [tex]\( h \)[/tex]. Here's how you do it step-by-step:
1. Square Both Sides to Eliminate the Square Root:
Start with the formula:
[tex]\[ v = \sqrt{2gh} \][/tex]
Square both sides:
[tex]\[ v^2 = 2gh \][/tex]
2. Rearrange the Equation to Solve for [tex]\( h \)[/tex]:
Divide both sides by [tex]\( 2g \)[/tex] to isolate [tex]\( h \)[/tex]:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
3. Substitute the Given Values:
Plug in the values for [tex]\( v \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[ h = \frac{4^2}{2 \times 32} \][/tex]
4. Calculate [tex]\( h \)[/tex]:
Simplify the equation:
- [tex]\( v^2 = 4^2 = 16 \)[/tex]
- [tex]\( 2 \times 32 = 64 \)[/tex]
Now compute:
[tex]\[ h = \frac{16}{64} \][/tex]
5. Simplify the Fraction:
[tex]\[ h = \frac{1}{4} = 0.25 \][/tex]
So, the hammer was dropped from a height of 0.25 feet above the ground. The correct answer is:
B. 0.25 feet